chem
posted by Lou .
The mass of the CuSO4 x 5H2O sample is 1.664 g. The mass of the recovered copper is 1.198 g.
Calculate the percent copper in pure copper sulfate pentahydrate(this is the theoretical percent copper). One mole of copper has a mass of 63.55 g, and by addition, one mole of CuSO4 x 5H2O has a mass of 249.5 g.

So what's the question? Percent Cu in the sample?
(1.198/1.664)*100 = ??
Theoretical yield =
(mass Cu/mass hydrate)*100 = (63.55/249.5)*100 = ?? 
The question is asking for percent of copper in the sample( so the first step shown above), but when I had to calculate the experimental percent I did the same thing, and i got 72%( is that the correct number of significant figures)?
I also wanted to know when I have to calculate the error of the % of copper what number do I use as true value and actual value? 
The mass of the CuSO4 x 5H2O sample is 1.664 g. The mass of the recovered copper is 1.198 g.
1. Compute the experimental percent by mass of copper in the sample:
1.1664/1.998x 100=71.81%
2. Calculate the percent copper in pure copper sulfate pentahydrate(this is the theoretical percent copper). One mole of copper has a mass of 63.55 g, and by addition, one mole of CuSO4 x 5H2O has a mass of 249.5 g:
63.55/249.5x100= 25.47%
3. In order to calculate the error of the % Cu and the percent error of the % Cu I know I am suppose to use 1.198 as the experimental value, but I am not sure what to use for the true value. 
Answered above.